Find one value of $x$ that is a solution to the equation: $(2x+3)^2-4(2x+3)-12=0$ $x=$
Explanation: We could solve for $x$ by expanding $(2x+3)^2$ and $-4(2x+3)$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that if we let ${p}={2x+3}$, we can rewrite the equation: $({2x+3})^2-4({2x+3})-12=0$ In particular, we can express it in the form: ${p}^2-4{p}-12=0$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2-4{p}-12&=0\\\\ ({p}-6)({p}+2)&=0\\\\ {p}=6\ &\text{or} \ \ {p}=-2 \end{aligned}$ Since ${p}={2x+3}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${2x+3}=6\ \ \ \text{or} \ \ \ {2x+3}=-2$ When we solve $2x+3=6$, we find that $x=\dfrac{3}{2}$. When we solve $2x+3=-2$, we find that $x=-\dfrac{5}{2}$. In conclusion, the two solutions of the equation $(2x+3)^2-4(2x+3)-12=0$ are $x=\dfrac{3}{2}$ and $x=-\dfrac{5}{2}$. [Is there another way to solve for x?]